For \(g \in W^{1,1}(0,T)\) , we can readily verify
\(J_t^{\alpha }\partial _t^{\alpha }g(t) = g(t) - g(0).\) 
Next we can prove
\(\Vert u(t)\Vert ^2 \in W^{1,1}(0,T) \quad \mbox{for $u_0 \in D(A)$}.\) 
Indeed,
\(& \partial _t\Vert u(t)\Vert ^2= \sum _{n=1}^{\infty } \langle u_0, \varphi _n\rangle ^2\partial _t(E_{\alpha ,1}(-\lambda _nt^{\alpha }))^2\\=& 2 \sum _{n=1}^{\infty } \langle u_0, \varphi _n\rangle ^2E_{\alpha ,1}(-\lambda _nt^{\alpha })(-\lambda _n)t^{\alpha -1}E_{\alpha ,\alpha }(-\lambda _nt^{\alpha }), \quad t>0.\) 

Dividing the summation \(\sum _{n=1}^{\infty }\)  into
\(\sum _{n=1}^m\)  and \(\sum _{n=m+1}^{\infty }\)  and using \(\lambda _n > 0\)  for
\(n\ge m+1\) , \(\lambda _n \le 0\)  for \(1\le n \le m\) , we can apply
Theorem 1.6 (p. 35) in [1]} for \(n\ge m+1\) , and
\(\vert E_{\alpha ,1}(-\lambda _nt^{\alpha })\vert \) ,
\(\vert E_{\alpha ,\alpha }(-\lambda _nt^{\alpha })\vert \le C_2\)  for
\(0\le t\le T\)  and \(1\le n \le m\) , we obtain
\(& \vert \partial _t\Vert u(t)\Vert ^2 \vert \le C_3 \sum _{n=1}^m \vert \lambda _n\vert \langle u_0,\varphi _n\rangle ^2t^{\alpha -1}+ C_3 \sum _{n=m+1}^{\infty } \vert \lambda _n\vert \langle u_0,\varphi _n\rangle ^2t^{\alpha -1}\left( \frac{1}{1+\lambda _nt^{\alpha }}\right)^2\\\le & C_4\left( \sum _{n=1}^{\infty } \vert \lambda _n\vert \langle u_0,\varphi _n\rangle ^2\right)t^{\alpha -1} \in L^1(0,T)\) 

by \(\alpha >0\) .
Thus the verification of (REF ) is complete.
By the Sobolev embedding, we remark that \(\Vert u(t)\Vert \in C[0,T]\) . Since \(J_t^{\alpha }g_1 \le J_t^{\alpha }g_2\)  in \((0,T)\)  for
\(g_1 \le g_2\)  in \((0,T)\) , we operate \(J_t^{\alpha }\)  to both sides of
(REF ). In terms of (REF ), we see
\(\Vert u(0)\Vert ^2 - \Vert u(T)\Vert ^2\le \frac{2}{\Gamma (\alpha )}\int ^T_0 (T-s)^{\alpha -1}\Vert u(s)\Vert \, \mathrm {d}s \, \Vert Au_0\Vert .\) 
Since \((T-s)^{\alpha -1} \in L^1(0,T)\)  as function in \(s\)  and
\(\Vert u(s)\Vert \in C[0,T]\)  by (REF ), the
mean value theorem for integral yields that there exists \(\xi =\xi (u_0)\in (0,T)\)  such that
\(\int ^T_0 (T-s)^{\alpha -1} \Vert u(s)\Vert \,\mathrm {d}s= \Vert u(\xi )\Vert \int ^T_0 (T-s)^{\alpha -1} \,\mathrm {d}s= \frac{T^{\alpha }}{\alpha }\Vert u(\xi )\Vert .\) 
Therefore, we obtain
\(\Vert u(0)\Vert ^{2} \le \Vert u(T)\Vert ^{2} + \frac{2\, T^\alpha }{\alpha \Gamma (\alpha )}\Vert u(\xi )\Vert  \Vert Au_0\Vert .\) 
By Theorem REF , we have
\(\Vert u(\xi )\Vert \le K\Vert u(0)\Vert ^{1-\frac{\xi }{T}}\Vert u(T)\Vert ^{\frac{\xi }{T}} \le KR^{1-\frac{\xi }{T}}\Vert u(T)\Vert ^{\frac{\xi }{T}}.\) 
Therefore, setting \(\theta := \frac{\xi }{2T}\) , we reach the conclusion.
This achieves the proof.
